How to generate two distinct numbers whose sum is equal to given number

Question

Hello guys ! I have a problem set in which I have to generate two distinct numbers whose sum is equal to the given number .

Suppose 8 is a number ,then I have to generate any two distinct numbers whose sum will be equal to 8.
Sample output.
Numebr:8
Two numbers are: 6 2
One more thing is there ,both numbers should not contain the digit 4.
Like it's should not be in this way----
Number :17
Two numbers are:13 4
Above output is wrong!!

So anyone one can help me in getting the logic

Answer 1

Just an idea I have in mind:

let x = 1, y = number - 1
loop while y greater than or equal to 1
if x + y equal to number
if x not equal to 4 and y not equal to 4
// we found a pair
x++
y--

But this way we may have multiple possibilities, I'm not sure which to choose if it happens. BTW is 4 an exception for any number of is it just for number 8? and how about 14 3 for number 17? 14 also has digit 4, so is 14 3 a valid pair for number 17?

(Edit)
I'm not even sure whether you want both number being not equals to 4 or neither number contains digit 4.

Answer 2

Pranshu Ranjan Welcome! 😊
I saw you solve, 👍

I just suggested, if it suits you 😉 Okay!

Have a nice coding friend! 🍻

Answer 3

Pranshu Ranjan
Will help if you check whether a number can be expressed as sum of two prime numbers?

Answer 4

Pranshu Ranjan,
Wow you're working on something great, I'll make sure to watch that video soon.

Thanks for sharing 👍

Answer 5

thank you friend for helping me I am really glad!!
i made some changes in the code with your logic and it's fine.
here is the code:-------------------------

import java.util.*;
public class Foregone{
public static void main(String[] args){
int n,x,y;
n=21;
x=1;
y=n-1;
while(y>=1){
if(x+y==n){
if(!(hasFour(x))&&!(hasFour(y))){
System.out.println("found pairs are "+y+" "+x);
break;
}
}
x++;
y--;
}

}
static boolean hasFour(int num) {
int rem;
while (num > 0) {
rem = num % 10;
if (rem == 4)
return true;
num = num / 10;
}
return false;
}
}

Answer 6

Pranshu Ranjan,
Good job my friend! you solved the problem. And now I understand clearly that you want neither x or y to have digit 4, great work! 👍

Answer 7

Yep! 4 should not be in the digits we get for any sum ,there is no exception for a particular number .
For sum 17 : 14 3
Both digits is okay but 14 contains the digit 4 ,that's why this answer would be wrong.
And I think you're logic is somewhere right !! I will look forward to implement it .
Thank for your response!!

Answer 8

It's good you have a solution.
Code on.

Answer 9

Ipang .
Thanks ! Actually this question was asked in one of the challenge of conducted by Google.

Hey friend Take time ,and watch my recent project also!
Presenting you Sam a digital assistant .its same like Microsoft crotana ,amazon alexa or say Jarvis is here .
Do watch and let me hear what you think about it.
Link is below---

https://youtu.be/jDsmymypNdo

Answer 10

Danijel
It's fine , I just wanted to get two distinct numbers and it's being solved .
Thanks for your response!

Answer 11

I won't give you the exact code, however try this approach.

Consider x is your target number.
nummax=x-1 nummin=1
Compare every value in between from nummin to nummax via stepping loop. Use logic to eliminate sum=x/2.

Answer 12

No need man! Everything is already done look above ,the code is written and is fine .
Thanks for your response and idea!!